Search: id:A026644
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%I A026644
%S A026644 1,2,4,10,20,42,84,170,340,682,1364,2730,5460,10922,21844,43690,
%T A026644 87380,174762,349524,699050,1398100,2796202,5592404,11184810,
%U A026644 22369620,44739242,89478484,178956970,357913940,715827882,1431655764
%N A026644 a(0) = 1, a(1) = 2, a(2) = 4; for n >= 3, a(n) = a(n-1) + 2*a(n-2) + 
               2.
%C A026644 Number of moves to solve Chinese rings puzzle.
%C A026644 a(n-1) (with a(0):=0) enumerates all sequences of length m=1,2,...,floor(n/
               2) with nonzero integer entries n_i satisfying sum |n_i| <= n-m. 
               Rephrasing K. A. Meissner's example in arXiv:gr-qc/0407052v1, p. 
               6. Example n=4: from length m=1: [1], [2], [3], each in 2 signed 
               versions; from m=2: [1,1] in 2^2=4 signed versions. Hence a(3)=a(4-1)=3*2+1*4=10.
%D A026644 Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 
               1991.
%D A026644 Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13-th 
               International Puzzle Party, Amsterdam, Aug 20 1993.
%H A026644 Lee Hae-hwang, <a href="a026644.html"> Illustration of initial terms 
               in terms of rosemary plants</a>
%F A026644 a(2k) = 2*a(2k-1), a(2k+1) = 2*a(2k)+2. - Peter Shor, Apr 11, 2002
%F A026644 For n>0: if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3. - Richard 
               Hess.
%F A026644 a(2n) = 2n-1 + Sum_{k = 0 to 2n-1} a(k), n>0; a(2n+1)= 2n+1 + Sum_{k=0 
               to 2n} a(k). - Lee Hae-hwang, Sep 17, 2002, corrected R. J. Mathar, 
               Oct 21 2008
%F A026644 a(n)=2n + 2*Sum_{k = 1 to n-2} a(k), n>0. - Lee Hae-hwang (mathmaniac(AT)empal.com), 
               Sep 19 2002, corrected R. J. Mathar, Oct 21 2008
%F A026644 G.f.: (1-x^2+2x^3)/((1-x)(1-x-2x^2)); a(n)=J(n+2)-1+0^n, where J(n)=A001045(n); 
               a(n)=2a(n-1)+a(n-2)-2a(n-3); a(n)=0^n+sum{k=0..n, (2-2*0^(n-k))J(k+1)}; 
               - Paul Barry (pbarry(AT)wit.ie), Oct 24 2007
%F A026644 a(n)=-1+(1/3)*(-1)^(n-1)+(8/3)*2^(n-1)+[C(2*n,n) mod 2], with n>=0 [From 
               Paolo P. Lava (ppl(AT)spl.at), Oct 03 2008]
%F A026644 a(n) = A052953(n+1)-2, n>0. [Moved from A020988, R. J. Mathar, (mathar(AT)strw.leidenuniv.nl), 
               Oct 21 2008]
%p A026644 f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi;
%Y A026644 a(n) = T(n, 0) + T(n, 1) + ... + T(n, n), T given by A026637.
%Y A026644 For n >= 1, equals twice A000975, also A001045 - 1.
%Y A026644 For n >= 1, a(n+1) = a(n) + 2*b(n+1) + 4*b(n), where b(k) = A001045(k).
%Y A026644 Sequence in context: A004647 A167030 A167193 this_sequence A026666 A121880 
               A094536
%Y A026644 Adjacent sequences: A026641 A026642 A026643 this_sequence A026645 A026646 
               A026647
%K A026644 nonn
%O A026644 0,2
%A A026644 Clark Kimberling (ck6(AT)evansville.edu)
%E A026644 Recurrence in definition line found by Lee Hae-hwang (mathmaniac(AT)empal.com), 
               Apr 03, 2002

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